Problem: Simplify the following expression: $y = \dfrac{6x^2- 17x- 3}{x - 3}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(6)}{(-3)} &=& -18 \\ {a} + {b} &=& &=& {-17} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-18$ and add them together. Remember, since $-18$ is negative, one of the factors must be negative. The factors that add up to ${-17}$ will be your ${a}$ and ${b}$ When ${a}$ is ${1}$ and ${b}$ is ${-18}$ $ \begin{eqnarray} {ab} &=& ({1})({-18}) &=& -18 \\ {a} + {b} &=& {1} + {-18} &=& -17 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({6}x^2 +{1}x) + ({-18}x {-3}) $ Factor out the common factors: $ x(6x + 1) - 3(6x + 1)$ Now factor out $(6x + 1)$ $ (6x + 1)(x - 3)$ The original expression can therefore be written: $ \dfrac{(6x + 1)(x - 3)}{x - 3}$ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ This leaves us with $6x + 1; x \neq 3$.